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Radda [10]
3 years ago
5

What is the relation of the sign of the charge on the charging body to the sign of the charge on the charged body when the charg

ing is by contact
Physics
1 answer:
galben [10]3 years ago
7 0

Charging by conduction involves the contact of a charged object to a neutral object. .

<u>Explanation:</u>

  • Suppose that a charged aluminum plate is touched to a neutral metal sphere, the neutral metal sphere becomes negatively charged because of the results of being contacted by the charged aluminum plate.
  • A charged metal sphere is touched to the highest plate of a neutral needle measuring instrument, the neutral measuring instrument becomes charged because of the results of being contacted by the metal sphere.
  • For example, a student standing on an insulating platform touches a negatively charged Van de Graaff generator, the neutrally charged student becomes charged. This is due to the contact made with the Van De Graaff generator which is negatively charged.

 

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Suppose astronomers built a 150-meter telescope. how much greater would its light-collecting area be than that of the 10-meter k
Dennis_Churaev [7]
Almost all telescopes have a circular mirror. The area of a circle is proportional to r^2 where r is the radius of the circle, the constant of proportionality being 4\pi
S =4\pi r^2
Therefore the area of 150 meter telescope would be 
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times bigger than the area of the smaller (10 meter) telescope.

6 0
2 years ago
A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is
GREYUIT [131]

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

W = \frac{1}{2}I\omega^2

now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

I = \frac{1}{2}(7 kg)(0.21^2)

I = 0.154 kg m^2

now from above equation we will have

39.6 = \frac{1}{2}(0.154)\omega^2

\omega = 22.67 rad/s

5 0
2 years ago
Read 2 more answers
(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the
frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.C=8\mu F=8\times 10^{-6} F

1\mu =10^{-6}

Q=48\mu C=48\times 10^{-6} C

We know that

V=\frac{Q}{C}

Using the formula

V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V

b.Q=192\mu C=192\times 10^{-6} C

V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V

8 0
3 years ago
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3 0
3 years ago
Tomas wants to paint his bedroom. He goes to the store and buys two cans of paint.
GuDViN [60]
C an inclined plate
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8 0
3 years ago
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